A screen beaming a real image of magnification $m_1$ formed by a convex lens is moved a distance x. The object is the moved until a new image of magnification $m_2$ is formed on the screen. The focal length of the lens is
Answer & explanation
Correct answer: option 1
In first case, $\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$ and $\frac{q}{p}=m_1$
$⇒1+m_1=\frac{q}{f}$ . . . (1)
In the second case $\frac{1}{q+x}+\frac{1}{p'}=\frac{1}{f}$
And $\frac{q+x}{p'}=m_2$
$⇒m_2=\frac{q+x}{f}$ . . . (2)
(1) and (2)
$⇒m_2-m_1=x/f⇒f=\frac{x}{m_2-m_1}$